What is the integral of cos^3(1/3 t) ?

The integral of cos^3(1/3 t) is 3sin(t/3)-sin^3(t/3)+C

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integral of cos^3(1/3 t)

\int cos^{3} (\frac{1}{3} t) d t

3 sin (\frac{t}{3}) - sin^{3} (\frac{t}{3}) + C

Solution steps

\int cos^{3} (\frac{1}{3} t) d t

Simplify

\frac{1}{3} t : \frac{t}{3}

= \int cos^{3} (\frac{t}{3}) d t

Apply u-substitution

: \int cos^{3} (u) \cdot 3 d u

= \int cos^{3} (u) \cdot 3 d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int cos^{3} (u) d u

Rewrite using trig identities

= 3 \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

: \int 1 - v^{2} d v

= 3 \cdot \int 1 - v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= 3 (v - \frac{v ^{3}}{3})

Substitute back

= 3 (sin (\frac{t}{3}) - \frac{sin ^{3} ( \frac{t}{3} )}{3})

Simplify

3 (sin (\frac{t}{3}) - \frac{sin ^{3} ( \frac{t}{3} )}{3}) : 3 sin (\frac{t}{3}) - sin^{3} (\frac{t}{3})

= 3 sin (\frac{t}{3}) - sin^{3} (\frac{t}{3})

Add a constant to the solution

= 3 sin (\frac{t}{3}) - sin^{3} (\frac{t}{3}) + C

\int x 8 - 4 x d x

\int 3 x (5 x + 2)^{3} d x

\int \frac{x ^{2}}{3 x + 18} d x

\int x 36 - x d x

What is the integral of cos^3(1/3 t) ?
The integral of cos^3(1/3 t) is 3sin(t/3)-sin^3(t/3)+C